Problem: A few families took a trip to an amusement park together. Tickets cost $$6.00$ each for adults and $$3.50$ each for kids, and the group paid $$52.00$ in total. There were $4$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Solution: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${6x+3.5y = 52}$ ${x = y-4}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-4}$ for $x$ in the first equation. ${6}{(y-4)}{+ 3.5y = 52}$ Simplify and solve for $y$ $ 6y-24 + 3.5y = 52 $ $ 9.5y-24 = 52 $ $ 9.5y = 76 $ $ y = \dfrac{76}{9.5} $ ${y = 8}$ Now that you know ${y = 8}$ , plug it back into ${x = y-4}$ to find $x$ ${x = }{(8)}{ - 4}$ ${x = 4}$ You can also plug ${y = 8}$ into ${6x+3.5y = 52}$ and get the same answer for $x$ ${6x + 3.5}{(8)}{= 52}$ ${x = 4}$ There were $4$ adults and $8$ kids.